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log (2x) - log (x+4) = log 5
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High School Teacher
log 2x - log (x+4) = log 5
We know that log a - log b = log a/b
==> log 2x/(x+4) = log5
==? 2x/(x+4) = 5
Multiply by (x+4):
==> 2x = 5*(x+4)
==> 2x = 5x + 20
==> 3x = -20
==> x = -20/3
Posted by hala718 on August 17, 2010 at 8:42 AM (Answer #1)
We'll impose the constraints of existence of logarithms:
The common interval of values that satisfies both constraints is (-0 , +inf.).
Now, we'll solve the equation. First, we'll add log (x+4) both sides:
log (2x) - log (x+4) + log (x+4) = log 5 + log (x+4)
log (2x) = log 5 + log (x+4)
Now, we'll use the product property of the logarithms:
log (2x) = log 5*(x+4)
Because the logarithms have matching bases, we'll use the one to one property:
2x = 5x + 20
We'll subtract 5x both sides:
2x-5x = 20
-3x = 20
We'll divide by -3:
x = -20/3 < 0
Since the solution is negative, is not admissible, so the equation has no solutions!
Posted by giorgiana1976 on August 17, 2010 at 9:45 PM (Answer #2)
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